PoAs in Mixed BGs

[Eques]

What happens when you are fighting a mixed battle group (eg muskets and pike) that produce different PoAs? Presumably you make separate rolls on a base by base basis.

But if so how do you calculate "lose 1 dice per/x"?

[ravenflight]

What happens when you are fighting a mixed battle group (eg muskets and pike) that produce different PoAs? Presumably you make separate rolls on a base by base basis.

But if so how do you calculate "lose 1 dice per/x"?

Generally in FoG:R the POA's for mixed BG's are the same. Like 'Protected shot, or pike in a mixed BG' (or words to that effect), but if you find a situation where that ISN'T the case then yes, you take dice per base in contact.

I had a situation where my Pike (armoured) had a POA advantage vs some Chinese bowmen. So, I had 2 dice at + and 4 dice at even. Didn't affect me, I'm still going for 4's. IF my opponent was disrupted (lose 1:3) then he would have had 6 dice becomes 4 dice. He'd proportionally lose the dice. I THINK that he'd end up with 2 dice going for 4's and 2 going for 5's. He'd be able to take the one going for 5's if he was to lose a third dice.

If he had already lost a base (a smart player would have taken a base facing the armoured pike - but lets assume he didn't) he would have 5 dice losing 1:3 means he loses 1 dice. The majority of the dice are going for 4's, so he loses one of them. Ends up exactly the same = 2 dice @ 4's 2 dice @ 5's. If he'd take. The base facing the pike he'd have had 3 dice @4's and 1 @ 5's

[rbodleyscott]

What happens when you are fighting a mixed battle group (eg muskets and pike) that produce different PoAs? Presumably you make separate rolls on a base by base basis.

But if so how do you calculate "lose 1 dice per/x"?

I had a situation where my Pike (armoured) had a POA advantage vs some Chinese bowmen. So, I had 2 dice at + and 4 dice at even. Didn't affect me, I'm still going for 4's. IF my opponent was disrupted (lose 1:3) then he would have had 6 dice becomes 4 dice. He'd proportionally lose the dice. I THINK that he'd end up with 2 dice going for 4's and 2 going for 5's. He'd be able to take the one going for 5's if he was to lose a third dice.

Actually it is 3 dice for 4s and 1 dice for 5s.

Think of it like this:

You start off with 4 dice for 4s and 2 dice for 5s.

You are losing 1 dice per 3 for being disrupted. So you have 6 dice and must lose 2.

Lose one of the dice for 4s and for the purpose of calculating the next lost dice deduct 3 dice from ones for 4s. (They have already paid their disruption penalty).

Remainder is 1 dice for 4s and 2 dice for 5s. It is therefore more in proportion to lose the second lost dice from the 2 dice for 5s than from the 1 remaining dice for 4s. Therefore you lose one dice for 5s.

So you have lost a total of 1 dice for 4s and 1 dice for 5s, leaving you with 3 dice for 4s and 1 dice for 5s.

[ravenflight]

Actually it is 3 dice for 4s and 1 dice for 5s.

Think of it like this:

You start off with 4 dice for 4s and 2 dice for 5s.

You are losing 1 dice per 3 for being disrupted. So you have 6 dice and must lose 2.

Lose one of the dice for 4s and for the purpose of calculating the next lost dice deduct 3 dice from ones for 4s. (They have already paid their disruption penalty).

Remainder is 1 dice for 4s and 2 dice for 5s. It is therefore more in proportion to lose the second lost dice from the 2 dice for 5s than from the 1 remaining dice for 4s. Therefore you lose one dice for 5s.

So you have lost a total of 1 dice for 4s and 1 dice for 5s, leaving you with 3 dice for 4s and 1 dice for 5s.

Thanks Richard. When you say it like that I realise that I would have calculated it like that 'on the table' just got all confunded in the heat of typing . (that's my excuse, and I'm sticking to it)